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Pie
4 years ago
6

When light hits a prism, the prism causes the light to ________?

Chemistry
1 answer:
guapka [62]4 years ago
7 0
Hello there,
The answer to your question is disperse

Hope this helps :))

~Top
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What is the equivalent to √-80
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Answer:

4i\sqrt{5}

Explanation:

i^{2}= -1

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Oxygen and hydrogen form the polyatomic hydroxide ion. What is its charge?
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negative

the chage on hydroxide ions os negative

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What was the purpose of letting the transformed cells sit in LB for a few minutes before spreading them onto the plates?
MissTica

Answer: D

Explanation:

6 0
3 years ago
an alloy composed of tin, lead, and cadmium is analyzed. the mole ratio of sn:pb is 2.73:1.00, and the mass ratio of pb:cd is 1.
nordsb [41]

This problem is describe the mole-ratio composition of an allow composed by tin, lead and cadmium. Ratios are given as Sn:Pb 2.73:1.00 and Pb:Cd is 1.78:1.00, and we are asked to calculate the mass percent compositon of Pb in the allow.

In this case, according to the given information, it turns out possible realize that the following number of moles are present in the alloy, according to the aforementioned ratios:

2.73mol Sn\\\\1.00molPb\\\\\frac{1.00molPb*1.00molCd}{1.78molPb}= 0.562molCd

Next, we calculate the masses by using each metal's atomic mass:

m_{Sn}=2.73mol*\frac{118.7g}{1mol}=324.05g\\\\ m_{Pb}=1.00mol*\frac{207.2g}{1mol}=207.2g\\\\m_{Cd}=0.562mol*\frac{112.4g}{1mol}=63.2g

Thus, the mass percent composition of each metal is shown below:

\%Sn=\frac{324.05g}{324.05g+207.2g+63.2g} *100\%=54.5\%\\\\\%Pb=\frac{207.2g}{324.05g+207.2g+63.2g} *100\%=34.9\%\\\\\%Cd=\frac{63.2}{324.05g+207.2g+63.2g} *100\%=10.6\%

So that of lead is 34.9 %.

  • brainly.com/question/19168996
  • brainly.com/question/2578569

Learn more:

8 0
3 years ago
You have a stock solution that is 226mg/mL and you need 10mL of a working solution that is 15mg/mL. What is the volume of stock
xenn [34]

Answer: 0.67 ml

Explanation:

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = concentration of stock solution = 226 mg/ml

V_1 = volume of stock solution = ?

C_2 = concentration of working solution= 15 mg/ml

V_2 = volume of working solution= 10 ml

Putting in the values we get:

226mg/ml\times V_1ml=15mg/ml\times 10ml

V_1=0.67ml

Thus volume of stock solution needed to dilute to have 10mL of working solution at the above concentration is 0.67 ml

6 0
3 years ago
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